$ \int_{-2}^4 \int_{2 - \sqrt{9 - (y - 1)^2}}^{2 + \sqrt{9 - (y - 1)^2}} dx \, dy$ Switch the bounds of the double integral. Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^4 \int_{1 - \sqrt{4 - x^2}}^{1 + \sqrt{4 - x^2}} dy \, dx$ (Choice B) B $ \int_{-1}^5 \int_{1 - \sqrt{9 - (x - 2)^2}}^{1 + \sqrt{9 - (x - 2)^2}} dy \, dx$ (Choice C) C $ \int_0^4 \int_{1 - \sqrt{4 - (x - 2)^2}}^{1 + \sqrt{4 - (x - 2)^2}} dy \, dx$ (Choice D) D $ \int_{-1}^5 \int_{1 - \sqrt{4 - (x - 1)^2}}^{1 + \sqrt{4 - (x - 1)^2}} dy \, dx$
The first step whenever we want to switch bounds is to sketch the region of integration that we're given. Here, we see $2 + \sqrt{9 - (y - 1)^2} < x < 2 - \sqrt{9 - (y - 1)^2}$ and $-2 < y < 4$. Therefore: ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${\llap{-}1}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${\llap{-}1}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ $y$ $x$ Because we're switching bounds to $dy \, dx$, we need to start with numeric bounds for $x$. We see that $-1 < x < 5$. Now we can define $y$ in terms of $x$. $1 - \sqrt{9 - (x - 2)^2} < y < 1 + \sqrt{9 - (x - 2)^2}$ We want to pay attention especially to how this $y$ bound works at the edge of the $x$ bound. For example, at $x = -1$, the $y$ bound makes $y = 1$ as expected. In conclusion, the double integral after switching bounds is: $ \int_{-1}^5 \int_{1 - \sqrt{9 - (x - 2)^2}}^{1 + \sqrt{9 - (x - 2)^2}} dy \, dx$